

Hi, all
Sorry for a dummy question.
I have some problems with the plot.
I use this code .
>> function xdot =F(x,t)
D =1.1;
r = 0.0984;
ka =3;
z =10;
umax=0.08;
km =3;
I0 =1;
xdot(1)= (x.*(D)).+((umax.*((I0.*2.717.^(ka.*x.*z))./((I0.*(2.717.^(ka.*x*z))).+km))).*x)
endfunction
>>
>>
>>
>> x0=9.8;
>> t= linspace(0,50,100);
>> quad("F",0,3);
I’ve got the xdot result.
But if I enter :plot(x,t)
error: 'x' undefined near line 1 column 6 comes out.
Can you tell me what the problem is ??
Best regards
Tae hoon lee


Hi, all
Sorry for a dummy question.
I have some problems with the plot.
I use this code .
>> function xdot =F(x,t)
D =1.1;
r = 0.0984;
ka =3;
z =10;
umax=0.08;
km =3;
I0 =1;
xdot(1)= (x.*(D)).+((umax.*((I0.*2.717.^(ka.*x.*z))./((I0.*(2.717.^(ka.*x*z))).+km))).*x)
endfunction
>>
>>
>>
>> x0=9.8;
>> t= linspace(0,50,100);
>> quad("F",0,3);
I’ve got the xdot result.
But if I enter :plot(x,t)
error: 'x' undefined near line 1 column 6 comes out.
Can you tell me what the problem is ??
Best regards
Tae hoon lee
You never defined a variable named x. What should x be?
Also, for what it's worth, the first input of the plot function is the independent variable, and the second is the dependent variable. I'm guessing you meant for t to be the independent variable.


Thanks, That’ right, my mistake. I want to see how the x changes over time. How should I define the x??? The x is the biomass, should I write the initial biomass value or make a function???
Please use "Reply all" so the entire mailing list can help. Also, in this mailing list we respond below the previous message rather than above.
Can you explain what it is you're trying to do? What is F? What variable are you trying to integrate it with respect to? I have no idea what biomass has to do with anything else.
I'm not sure if it's what you were asking, but when you call the plot function, the arguments should be arrays with the same length. For example,
n=100;
x = linspace(0,2*pi,n); for j=1:n y(j) = x(j)*sin(x(j)); end plot(x,y)


Dear. Brett My purpose is to see how the x(biomass) changes over time. The equation for biomass is x=x0(initial biomass)*exp(u(growth rate)D(dilution rate)*t). And the “u”(growth rate) can be substitute to u =I/(I+km). The “I”(light intensity) can be substitute I=I0*exp(x(biomass)*z(distance)*ka(absorbance coefficient))
The integration of the equation dx/dt=(uD)x . If I binomialize this, I get ∫(1/x)dx(from x0 to x)= ∫(uD)dt(from t0 to t) . And the result is x=x0*exp(uD)*t.
But when you substitute the “u” with I/(I+ka) The right side "∫(uD)dt" can’t be integrated with respect to time because the light intensity changes over time.
So I first tried with eq . x=x0*exp((uD)*t) the ode45 to see how the x changes over time. What I’m doing now is to make a direct integration method to compare it with the numerical approach(Ode45).
Thanks, That’ right, my mistake. I want to see how the x changes over time. How should I define the x??? The x is the biomass, should I write the initial biomass value or make a function???
Please use "Reply all" so the entire mailing list can help. Also, in this mailing list we respond below the previous message rather than above.
Can you explain what it is you're trying to do? What is F? What variable are you trying to integrate it with respect to? I have no idea what biomass has to do with anything else.
I'm not sure if it's what you were asking, but when you call the plot function, the arguments should be arrays with the same length. For example,
n=100;
x = linspace(0,2*pi,n); for j=1:n y(j) = x(j)*sin(x(j)); end plot(x,y)


Your "direct integration" means an analytical formula of X(t) as a function
of t, with several constant parameters, like ka, I0, x0...etc. Right?
If that is the case, then you may construct your function as:
Function x = F(t)
%define constant parameters
% x = your analytical formula
end
then depends what range of t you want, you may assign:
t = linspace(0,t_end,100);
x = F(t);
plot(x,y);

Sent from: https://octave.1599824.n4.nabble.com/OctaveGeneralf1599825.html

